Answer
$P \approx 3868.904$
Work Step by Step
$$\eqalign{
& {\text{The rate of change of }}P{\text{ is proportional to }}P,{\text{ then }} \cr
& \frac{{dP}}{{dt}} = kP \cr
& P' = kP \cr
& {\text{Solving for }P \text{ we obtain}} \cr
& P = C{e^{kt}}{\text{, }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{When }}t = 0,{\text{ }}P = 5000,{\text{ then}} \cr
& 5000 = C{e^{k\left( 0 \right)}} \cr
& {\text{Solve for }}C \cr
& 5000 = C\left( 1 \right) \cr
& C = 5000 \cr
& \cr
& {\text{Substitute }}C = 5000\,{\text{into }}\left( {\bf{1}} \right){\text{ }} \cr
& P = 5000{e^{kt}}{\text{, }}\left( {\bf{2}} \right) \cr
& \cr
& {\text{When }}t = 1,{\text{ }}P = 4750,{\text{ then}} \cr
& 4750 = 5000{e^{k\left( 1 \right)}} \cr
& {\text{solve for }}k \cr
& {e^k} = \frac{{4750}}{{5000}} \cr
& {e^k} = \frac{{19}}{{20}} \cr
& k = \ln \left( {\frac{{19}}{{20}}} \right) \cr
& \cr
& {\text{Substitute }}k = \ln \left( {\frac{{19}}{{20}}} \right){\text{ into the equation }}\left( {\bf{2}} \right) \cr
& P = 5000{e^{\ln \left( {\frac{{19}}{{20}}} \right)t}} \cr
& \cr
& {\text{Calculate }}P{\text{ when }}t = 5 \cr
& P = 5000{e^{\ln \left( {\frac{{19}}{{20}}} \right)\left( 5 \right)}} \cr
& P \approx 3868.904 \cr} $$
